根据组和另一个数据框删除 R 数据框中的行答案

【问题标题】：remove rows in R data frame based on group and another data frame根据组和另一个数据框删除 R 数据框中的行
【发布时间】：2019-02-04 11:39:24
【问题描述】：

有两个数据框

dat1 <- data.frame(group= c(11,11,12,12,13,13,14,14,15,15,16,16,17,17,17,18,18,18),name= c("A","B","C","D","E","F","G","H","I","J","A","B","E","F","W","A","B","V"))

dat2 <- data.frame(ID=c(1,1,2,2,3,3),name =c("A","B","E","F","X","Y"))

第二个数据框具有按 ID 列分组的两个值的组合。并且基于第二个数据帧 (dat2) ，如果特定的组组合存在于 dat2 中，则需要删除第一个数据帧 (dat1) 中的行。

例如：如果“A”和“B”都存在于 dat1 中，则应将其删除。

因此，期望的输出是

desiredat <- data.frame(group= c(12,12,13,13,15,15),name= c("C","D","G","H","I","J"))

在 R 中寻找实现相同目标的方法。

【问题讨论】：

检查你的desiredat 我认为有一些值遗漏了

标签： r dataframe

【解决方案1】：

这可以通过反连接来解决。但是，我们需要确定哪些组 ID group 必须从 dat1 中删除。

library(data.table)
# count names per ID
setDT(dat2)[, n.id := .N, by = ID]
# identify groups to remove by joining and ... 
groups_to_remove <- dat2[setDT(dat1), on = "name", nomatch = 0L][
  # ... check which groups have a match with the complete set of names
  , which(n.id == .N), by = .(ID, group)]
# anti join
dat1[!groups_to_remove, on = "group"]

   group name
1:    12    C
2:    12    D
3:    14    G
4:    14    H
5:    15    I
6:    15    J
7:    19    A
8:    19    X

第 19 组没有被删除，因为名称“A”和“X”属于 dat2 中的不同 ID。

更精简的方法使用all()，而不是计算唯一名称：

library(data.table)
setDT(dat1)
setDT(dat2)
groups_to_remove <- dat1[dat2, on = "name"][, which(all(ID == ID[1])), by = group]
dat1[!groups_to_remove, on = "group"]

   group name
1:    12    C
2:    12    D
3:    14    G
4:    14    H
5:    15    I
6:    15    J
7:    19    A
8:    19    X

同上dplyr语法：

library(dplyr)
dat2 %>% 
  left_join(dat1, by = "name") %>% 
  group_by(group) %>% 
  summarise(all_have_same_id = all(ID == ID[1L])) %>% 
  filter(all_have_same_id) %>% 
  anti_join(dat1, ., by = "group")

  group name
1    12    C
2    12    D
3    14    G
4    14    H
5    15    I
6    15    J
7    19    A
8    19    X
Warning message:
Column `name` joining factors with different levels, coercing to character vector

数据

OP 提供的样本数据集dat1 由以下组组成其中，dat2 的 ID 中仅包含一个名称。因此，我添加了这个用例（作为第 19 组）：

dat1 <- data.frame(
  group= c(11,11,12,12,13,13,14,14,15,15,16,16,17,17,17,18,18,18,19,19),
  name= c("A","B","C","D","E","F","G","H","I","J","A","B","E","F","W","A","B","V","A","X"))

dat2 <- data.frame(ID=c(1,1,2,2,3,3),name =c("A","B","E","F","X","Y"))

【讨论】：

【解决方案2】：

像这样...？

dat1[dat1$name %in% setdiff(dat1$name, dat2$name), ]
3     12    C
4     12    D
7     14    G
8     14    H
9     15    I
10    15    J
15    17    W
18    18    V

【讨论】：